Problem 337 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \left.\lim _{x \rightarrow 2} \frac{\ln \left(x^{2}-3\right)}{x^{2}+3 x-10} \text { \\{Ans. } \frac{4}{7}\right\\} $$

Step-by-Step Solution

Answer

The short answer to the given limit problem is: \(\lim_{x \rightarrow 2} \frac{\ln\left(x^2 - 3\right)}{x^2 + 3x - 10} = \frac{4}{7}\)

Step 1: Identify the lim of the function

We are given the limit: \[\lim_{x \rightarrow 2} \frac{\ln\left(x^2 - 3\right)}{x^2 + 3x - 10}\]

Step 2: Factor the denominator

Factor the quadratic equation in the denominator: \(x^2 + 3x - 10 = (x + 5)(x - 2)\)

Step 3: Rewrite the function

Now we rewrite the function with the factored denominator: \(f(x) = \frac{\ln\left(x^2 - 3\right)}{(x + 5)(x - 2)}\)

Step 4: Apply L'Hopital's Rule

Now, since we have an indeterminate form \(\frac{0}{0}\) when x = 2, as both numerator and denominator approach zero, we can apply L'Hopital's Rule. Differentiate the numerator and denominator separately with respect to x to find their derivatives: \(f'(x) = \frac{\frac{d}{dx}\left(\ln\left(x^2 - 3\right)\right)}{\frac{d}{dx}\left((x + 5)(x - 2)\right)}\)

Step 5: Differentiate the numerator

Differentiate the natural logarithm function using the chain rule: \(\frac{d}{dx}\left(\ln\left(x^2 - 3\right)\right) = \frac{1}{x^2 - 3} \cdot 2x = \frac{2x}{x^2 - 3}\)

Step 6: Differentiate the denominator

Differentiate the factored quadratic denominator using the sum/difference rule: \(\frac{d}{dx}\left((x + 5)(x - 2)\right) = (x + 5) + (x - 2) = 2x + 3\)

Step 7: Calculate the new limit

Now we can find the new limit with the derivatives: \(\lim_{x \rightarrow 2} \frac{\frac{2x}{x^2 - 3}}{2x + 3}\)

Step 8: Plug in x = 2

Evaluate the limit by plugging in x = 2: \(\frac{\frac{2(2)}{(2)^2 - 3}}{2(2) + 3} = \frac{\frac{4}{1}}{7} = \frac{4}{7}\) The limit of the given function as x approaches 2 is indeed \(\frac{4}{7}\).

Key Concepts

Limit Calculation

Limit calculation is a crucial concept in calculus. It helps us understand the behavior of functions as they get closer to a particular point. In this exercise, we're assessing the limit of \[\lim_{x \rightarrow 2} \frac{\ln\left(x^2 - 3\right)}{x^2 + 3x - 10}\]This asks us what value the function approaches as \(x\) approaches 2. It's like asking, "Where are we headed as we get really close to 2 on the x-axis?" Often, direct substitution works, but here the direct attempt resulted in an indeterminate form. To tackle such situations, we need approaches that go beyond merely plugging in numbers.

Derivative

Derivatives measure how a function changes as its input changes. This rate of change can be visualized as the slope of the function’s graph at any point.In the context of L'Hopital's Rule, which we see applied here, a derivative allows us to find new expressions for our limit calculation. When both top and bottom parts of the original fraction become zero (indeterminate form), we must differentiate both the numerator and the denominator. For instance, given \(\ln(x^2 - 3)\), we use the chain rule to derive its derivative: \[\frac{d}{dx}\left(\ln(x^2 - 3)\right) = \frac{2x}{x^2 - 3}\].Likewise, the derivative of the denominator, a polynomial, is found by the sum/difference rule, resulting in \(2x + 3\). These derivatives are then used to compute the new limit.

Indeterminate Forms

When evaluating limits, indeterminate forms such as \(\frac{0}{0}\) can arise. In simple terms, an indeterminate form occurs when plugged-in values lead to non-defined results.For this problem, the form \(\frac{0}{0}\) was encountered as both the numerator and the denominator equaled zero at \(x = 2\). The indeterminate form signifies that more work is needed to evaluate the limit. L'Hopital's Rule provides the tool to deal with these cases by replacing the original fraction with \(\frac{f'(x)}{g'(x)}\). This substitution gives a determinate form, facilitating easier calculation of the limit. Remember: the rule applies when both numerator and denominator tend to zero or infinity at the same point.

Factorization

Factorization is about breaking down an expression into a product of simpler expressions. In our problem’s original denominator, \(x^2 + 3x - 10\), we factor it before verifying the indeterminate form: \((x + 5)(x - 2)\). Factorization can simplify complex expressions and is often the first step in solving problems involving polynomials. By transforming the denominator, we can clearly identify points where it becomes zero, allowing us to determine points of discontinuity or singularity, aiding in the application of L'Hopital's Rule. Thus, factorization remains a valuable tool in the calculus toolbox for clearer insights and easier manipulation of algebraic expressions.